Three dimensional puzzle

ABSTRACT

A three-dimensional puzzle which forms a regular polyhedron and has not conventionally existed is realized. In addition, a three-dimensional puzzle which realizes a Fedrov space filling solid and has not conventionally existed is realized. According to the invention, a three-dimensional puzzle is provided having a regular polyhedron consisting of a plurality of convex polyhedrons which fill an interior of the regular polyhedron comprising the plurality of convex polyhedrons having a plurality of a pair of convex polyhedrons in a mirroring image relationship, wherein the plurality of convex polyhedrons are indivisible into two or more congruent shaped polyhedrons. In addition, the plurality of convex polyhedrons may be four convex polyhedrons and include three pairs of convex polyhedrons in a mirroring image relationship. Further the plurality of convex polyhedrons may be five convex polyhedrons and include four pairs of convex polyhedrons in a mirroring image relationship.

CROSS REFERENCE TO RELATED APPLICATIONS

This application is based upon and claims the benefit of priority fromthe prior Japanese Patent Application No. 2008-156057, filed on Jun. 14,2008, Japanese Patent Application No. 2008-268221, filed on Oct. 17,2008, Japanese Patent Application No. 2008-277198, filed on Oct. 28,2008, and PCT Application No. PCT/JP2009/060763, filed on Jun. 12, 2009,the entire contents of which are incorporated herein by reference.

BACKGROUND OF THE INVENTION

1. Technical Field

The present invention is related to a three-dimensional puzzle. Inparticular, the present invention is related to a regular polyhedronpuzzle which includes a plurality of convex polyhedrons which fill aninterior, and a three-dimensional puzzle which can realize a Fedorovspace-filling three dimensional solid.

2. Description of the Related Art

A regular polyhedron puzzle is known in which the regular polyhedron isdivided into several convex polyhedrons. The convex polyhedrons whichform this conventional regular polyhedron puzzle do not have particularregularity or characteristics, and are appropriately divided in order toadjust the difficulty and complexity of the puzzle (e.g. Japan Laid OpenUtility Model Publication No. S63-200867, Japan Laid Open PatentPublication No. S58-049168, Japan Registered Utility Model ApplicationNo. 3107739, International Laid Open Pamphlet No. 2006/075666 and JapanLaid Open Patent Publication No. H03-155891).

BRIEF SUMMARY OF THE INVENTION

The present invention forms a regular polyhedron and realizes athree-dimensional puzzle which has not conventionally existed. Inaddition, it is possible to realize a Fedrov space filling solid andrealize a three-dimensional puzzle which has not conventionally existed.

According to one embodiment of the present invention, athree-dimensional puzzle is provided including four types of convexpolyhedrons from which a regular tetrahedron, a cube, a regularoctahedron, a regular dodecahedron or a regular icosahedron are formed,wherein three types among the four types of convex polyhedrons eachhaving a pair of convex polyhedrons in a mirroring image relationship,the four types of convex polyhedrons are indivisible into two or morecongruent shaped polyhedrons, and the regular tetrahedron, the cube, theregular octahedron, the regular dodecahedron and the regular icosahedronare formed using only the four types of convex polyhedrons so that theinterior of the regular tetrahedron, the cube, the regular octahedron,the regular dodecahedron and the regular icosahedron are filled.

In addition, according to one embodiment of the present invention athree-dimensional puzzle is provided including five types of convexpolyhedrons from which a regular tetrahedron, a cube, a regularoctahedron, a regular dodecahedron or a regular icosahedron are formed,wherein four types among the five types of convex polyhedrons eachhaving a pair of convex polyhedrons in a mirroring image relationship,the five types of convex polyhedrons are indivisible into two or morecongruent shaped polyhedrons, and the regular tetrahedron, the cube, theregular octahedron, the regular dodecahedron and the regular icosahedronare formed using only the five types of convex polyhedrons so that theinterior of the regular tetrahedron, the cube, the regular octahedron,the regular dodecahedron and the regular icosahedron are filled.

According to one embodiment of the present invention, a Fedorovspace-filling three-dimensional puzzle is provided including a pluralityof first convex polyhedrons and a plurality of second convex polyhedronswhich are in a mirroring relationship with the first convex polyhedronsfrom which the three-dimensional puzzle is formed, wherein the pluralityof first convex polyhedrons and the plurality of second convexpolyhedrons each are indivisible into two or more congruent shapedpolyhedrons, and the plurality of first convex polyhedrons and theplurality of second convex polyhedrons form all the Fedorovspace-filling three dimensional solids by filling the interior of allthe Fedorov space-filling three dimensional solids.

The Fedorov space-filling three-dimensional puzzle may also include anelongated rhombic dodecahedron which is formed using the plurality offirst convex polyhedrons and the plurality of second convex polyhedronsincludes a truncated octahedron, a parallel hexahedron, a skewedhexagonal prism and a rhombic dodecahedron.

BRIEF DESCRIPTION OF THE DRAWINGS

FIG. 1 is a diagram which shows a three-dimensional puzzle (regulartetrahedron) 100 related to one embodiment of the present invention;

FIG. 2( a) is a diagram which shows a cube 200, FIG. 2( b) is a diagramwhich shows a triangular pyramid 201 including a peak B which has beencut away;

FIG. 3( a) is a diagram which shows the appearance of forming aquadrangular pyramid 301 from four triangular pyramids 201, FIG. 3( b)is a diagram which shows the appearance of forming a regular octahedron300 from two quadrangular pyramids 301;

FIG. 4( a) is a diagram which shows three points Q, T, U respectivelyplaced on the three edges of the triangular pyramid 201, FIG. 4( b) is adiagram which shows a triangular pyramid 203 including a peak A whichhas been cut away from the triangular pyramid 201;

FIG. 5( a) and (b) are diagrams which show the triangular pyramid 203including the peaks C, F which has been cut away, FIG. 5( c) is adiagram which shows a regular icosahedron 400 formed from four pairs ofa heptahedron 207 and a mirror image symmetrical heptahedron 209;

FIG. 6( a) is a diagram which shows a regular dodecahedron 500, FIG. 6(b) is a diagram which shows a cut away through a plane which passesthrough the peaks A, B, C, D and a plane which passes through the peaksA, B, E, F, FIG. 6( c) shows the appearance of the cube 200 which isleft at the center of the cut away;

FIG. 7( a) is a diagram which shows a view of the regular tetrahedron100 again, FIG. 7( b) is a diagram which shows the regular tetrahedron100 cut away through a plane which passes through the center points I,J, K, L on the four edges AF, FC, CH, and HA, FIG. 7( c) is a diagram ofan extracted prism 101, FIG. 7( d) is a diagram of a cut away through aplane which passes through three points J, L, M when M is the centerpoint of the edge FH, FIG. 7( e) is a diagram in which a pentahedron 103is divided into two congruent tetrahedrons 110 and 111;

FIG. 8 is a diagram which shows a development of a piece 110 which isthe shape of an atom α which forms the three-dimensional puzzle relatedto one embodiment of the present invention;

FIG. 9 is a diagram which shows a view of an equihepta 207 again, FIG.9( b) is a diagram which shows the appearance of the equihepta dividedinto three solids;

FIG. 10 is a development of a piece 210 which is the shape of an atom γwhich forms the three-dimensional puzzle related to one embodiment ofthe present invention;

FIG. 11 is a development of a piece 203 which is the shape of an atom γwhich forms the three-dimensional puzzle related to one embodiment ofthe present invention;

FIG. 12( a) is a diagram which shows a roof 501 related to oneembodiment of the present invention again, FIG. 12( b) shows a piece 510which is the shape of an atom δ which forms the three-dimensional puzzlerelated to one embodiment of the present invention;

FIG. 13 shows the piece 510 which is the shape of an atom δ which formsthe three-dimensional puzzle related to one embodiment of the presentinvention;

FIG. 14 is a diagram which shows a cube three-dimensional puzzle 200related to one embodiment of the present invention;

FIG. 15 is a diagram of parts 201 of a triangular pyramid related to oneembodiment of the present invention which is formed using parts 207comprised of a piece 210 which is the shape of the atom β and a piece203 which is the shape of the atom γ;

FIG. 16 is a diagram of parts 207 related to one embodiment of thepresent invention which is formed using a piece 210 which is the shapeof three atoms β;

FIG. 17 is a diagram which shows a regular octahedron three-dimensionalpuzzle 300 related to one embodiment of the present invention;

FIG. 18 is a diagram which shows a regular dodecahedronthree-dimensional puzzle 500 related to one embodiment of the presentinvention;

FIG. 19 is a diagram which shows a regular icosahedron three-dimensionalpuzzle 400 related to one embodiment of the present invention;

FIG. 20( a) is a diagram which shows the appearance of the roof 501which is reversed so as to fill the interior of the cube 200, FIG. 20(b) is a diagram which shows a hexahedron 601 related to one embodimentof the present invention, FIG. 20( c) shows a development of thehexahedron 601, FIG. 20( d) is a diagram which shows the appearance whenthe interior of the cube 200 is filled with the two mirror imagesymmetrical hexahedrons 601 and 602;

FIG. 21 is a development of a piece 610 which is the shape of an atom εrelated to one embodiment of the present invention;

FIG. 22( a) is a view of the piece 110 which is the shape of the atom αagain, FIG. 22( b) is a diagram which shows the piece 110 divided into atetrahedron 121 and a tetrahedron 123, FIG. 22( c) is a diagram whichshows a piece 151 which is the shape of an atom ξ and a piece 152 whichis the shape of an atom ξ′ related to one embodiment of the presentinvention, FIG. 22( d) is a diagram which shows a quadrangular pyramidformed by gluing a tetrahedron 123 and a tetrahedron 124;

FIG. 23 is a development of the piece 151 which is the shape of the atomξ related to one embodiment of the present invention;

FIG. 24 is a diagram which shows the appearance of an end of the piece203 which is the shape of the atom γ which is divided into a piece 800which is the shape of the atom θ and a piece 900 which is the shape ofthe atom η;

FIG. 25 is a development of the piece 800 which is the shape of the atomθ related to one embodiment of the present invention;

FIG. 26 is a development of the piece 900 which is the shape of the atomη related to one embodiment of the present invention;

FIG. 27 is a diagram which shows a piece 700 which is the shape of apolyhedron corresponding to two atoms α related to an embodiment of thepresent invention;

FIG. 28 is a diagram which shows a regular tetrahedron three-dimensionalpuzzle related to one embodiment of the present invention;

FIGS. 29( a) to (e) are diagrams of five types of Fedorov space-fillingthree-dimensional puzzles related to one embodiment of the presentinvention. FIG. 29( a) shows a parallel hexahedron, FIG. 29( b) shows arhombic dodecahedron, FIG. 29( c) shows a skewed hexagonal prism, FIG.29( d) shows an elongated rhombic dodecahedron, FIG. 29( e) shows atruncated octahedron;

FIGS. 30( a) to (e) are exemplary diagrams which show the appearancewhere space is filled by stacking of translates of five types of Fedorovspace-filling three-dimensional puzzles. FIG. 30( a) shows theappearance of space filling by a parallel hexahedron, FIG. 30( b) showsthe appearance of space filling by a rhombic dodecahedron, FIG. 30( c)shows the appearance of space filling by a skewed hexagonal prism, FIG.30( d) shows the appearance of space filling by a elongated rhombicdodecahedron, FIG. 30( e) shows the appearance of space filling by atruncated octahedron;

FIG. 31( a) is a diagram which shows a cube 1200, FIG. 31( b) shows thecube divided into six congruent quadrangular pyramids 1201;

FIG. 32( a) is a diagram which shows one of the quadrangular pyramids1201 FIG. 32( b) shows one of the quadrangular pyramids 1201 dividedinto right tetrahedrons 1203 by two perpendicular planes, FIG. 32( c) isa diagram which shows a right tetrahedron 1203;

FIG. 33( a) shows an exemplary view where the quadrangular pyramids 1201are glued to each side of the cube 1200, FIG. 33( b) is a diagram whichshows a rhombic dodecahedron 1300;

FIG. 34( a) shows a sphenoid 1100 in which two right tetrahedrons 1203are glued to the bottom side of right angled isosceles triangles, FIG.34( b) shows a triangular prism 1401 in which three sphenoids 1100 arecombined together, FIG. 34( c) shows a skewed hexagonal prism 1400;

FIG. 35( a) shows a rhombic dodecahedron 1300, FIG. 35( b) is anexemplary diagram which shows the formation of an elongated rhombicdodecahedron 1500 by the rhombic dodecahedron 1300 and a helmet-shapedpolyhedron 1501;

FIG. 36( a) is a diagram which shows the sphenoid 1100 related to oneembodiment of the present invention, FIG. 36( b) shows the sphenoid 1100divided into c-squadrons 1105;

FIG. 37( a) is an enneahedron of a diamond 1601 in which fourc-squadrons 1105 are combined together, FIG. 37( b) is an exemplarydiagram of a truncated octahedron (1600) formed by arranging six of theenneahedrons 1601;

FIG. 38( a) is a diagram which shows a c-squadron 1105 related to oneembodiment of the present invention, FIG. 38( b) is an exemplary diagramwhich shows the c-squadron 1105 divided into σ 1101 and σ′ 1103, FIG.38( c) is an exemplary diagram which shows the formation of a righttetrahedron 1203 using two each of σ 1101 and σ′ 1103;

FIG. 39 is a development of σ 1101 and σ′ 1103 related to one embodimentof the present invention;

FIG. 40 is an exemplary diagram which shows the formation of ahexahedron 1701 using σ 1101 and σ′ 1103 related to one embodiment ofthe present invention;

FIG. 41( a) is an upper side diagram which shows the formation of apolyhedron 1703 using three hexahedrons 1701, FIG. 41( b) is a bottomside diagram of the polyhedron 1703;

FIG. 42( a) is an exemplary diagram which shows the formation of a cube1705 with an open hole using the polyhedron 1703, FIG. 42( b) is adiagram which shows a combination of the cube 1705 with open hole andhalf of the truncated octahedron 1603, FIG. 42( c) is a diagram whichshows half of a cube formed by the cube 1705 with open hole and half ofthe truncated octahedron 1603, FIG. 42( d) is a diagram which shows thecube 1700 which is formed;

FIG. 43( a) is a diagram which shows half of a rhombic dodecahedron1301, FIG. 43( b) is a diagram which shows a helmet-shaped polyhedron1501, FIG. 43( c) is a diagram which shows a polyhedron 1503 cut awayfrom the helmet-shaped polyhedron 1501, FIG. 43( d) is a diagram whichshows the formation of a skewed hexagonal prism 1420 from the polyhedron1503 and half of the rhombic dodecahedron 1301, FIG. 43( e) is a diagramwhich shows the skewed hexagonal prism 1420;

FIG. 44 shows an arrangement of a puzzle 1651 in which a truncatedoctahedron three-dimensional puzzle related to one embodiment of thepresent invention is halved, FIG. 44( a) shows a side face view, FIG.44( b) shows a side face view, FIG. 44( c) shows an upper face view,FIG. 44( d) shows a bottom face view;

FIG. 45 shows an arrangement of a puzzle 1650 in which a truncatedoctahedron three-dimensional puzzle related to one embodiment of thepresent invention is halved, FIG. 45( a) shows a side face view, FIG.45( b) shows a side face view, FIG. 45( c) shows a bottom face view;

FIG. 46 is a diagram which shows an arrangement of a cubethree-dimensional puzzle related to one embodiment of the presentinvention, FIG. 46( a) shows a bottom face view of a puzzle 1651 inwhich a truncated octahedron three-dimensional puzzle is halved, FIG.46( b) shows a puzzle 1751 formed by cutting a cube with an open hole,FIG. 46( c) shows half of a cube puzzle 1753 in which the puzzle 1651which is half of a truncated octahedron three-dimensional puzzle, isglued to the puzzle 1751 formed by cutting a cube with an open hole,FIG. 46( d) shows a cube three-dimensional puzzle 1750;

FIG. 47 is a diagram which shows an arrangement of a rhombicdodecahedron three-dimensional puzzle related to one embodiment of thepresent invention, FIG. 47( a) shows half of a cube puzzle 1753, FIG.47( b) shows a puzzle 1353 where half of a cube is removed from a puzzle1351 which is half of a rhombic dodecahedron, FIG. 47( c) shows thepuzzle 1351 which is half of a rhombic dodecahedron, FIG. 47( d) shows aside face view of a rhombic dodecahedron puzzle 1350, FIG. 47.(e) showsan upper side view of the rhombic dodecahedron puzzle 1350;

FIG. 48 is a diagram which shows an arrangement of a skewed hexagonalprism three-dimensional puzzle related to one embodiment of the presentinvention, FIG. 48( a) shows a side face view of a puzzle 1553 which hasbeen cut away from a helmet-shaped polyhedron puzzle 1551, FIG. 48( b)shows a front face view of FIG. 48( a), FIG. 48( c) shows a back faceview of FIG. 48( a), FIG. 48( d) shows an exemplary diagram of theformation of a skewed hexagonal prism puzzle 1450 by the puzzle 1553which is cut away from the helmet-shaped polyhedron 1551, and the puzzle1351 which is half of the rhombic dodecahedron 1350, FIG. 48( e) showsan upper face view of the skewed hexagonal prism puzzle 1450, FIG. 48(f) shows a side face view of the skewed hexagonal prism puzzle 1450; and

FIG. 49 is a diagram which shows an arrangement of an elongated rhombicdodecahedron three-dimensional puzzle related to one embodiment of thepresent invention, FIG. 49( a) shows a side face view of thehelmet-shaped polyhedron puzzle 1551, FIG. 49( b) shows a front faceview of the helmet-shaped polyhedron puzzle 1551, FIG. 49( c) shows aback face view of the helmet-shaped polyhedron puzzle 1551, FIG. 49( d)shows an exemplary diagram of the formation of an elongated rhombicdodecahedron puzzle 1550 by the helmet-shaped polyhedron puzzle 1551 andthe rhombic dodecahedron puzzle 1350, FIG. 49( e) shows an upper faceview of the elongated rhombic dodecahedron puzzle 1550, FIG. 49( f)shows a side face view of the elongated rhombic dodecahedron puzzle1550.

EXPLANATION OF THE REFERENCE SYMBOLS

-   100 regular tetrahedron-   101 prism-   103 pentahedron-   110 atom α-   111 atom α′-   121 tetrahedron LJKW-   123 tetrahedron LHKW-   124 mirror image of tetrahedron 123-   151 atom ξ-   152 atom ξ-   200 cube-   201 regular triangular pyramid-   203 golden tetra (atom γ)-   204 atom γ′-   205 polyhedron except golden tetra from regular triangular pyramid-   207 equihepta-   209 mirror image of equihepta-   210 atom β-   211 atom β′-   300 regular octahedron-   301 quadrangular pyramid-   400 regular icosahedron-   500 regular dodecahedron-   501 roof-   510 atom δ-   601 hexahedron-   602 mirror image of hexahedron 601-   610 atom ε-   611 atom ε′-   700 α2-   701 quadrangular pyramid LHKWK′-   800 atom θ-   801 atom θ′-   900 atom η-   901 atom η′-   1100 sphenoid-   1101 σ-   1103 σ′-   1105 c-squadron-   1200 cube-   1201 quadrangular pyramid-   1203 triangular pyramid (right tetrahedron)-   1300 rhombic dodecahedron-   1301 polyhedron which is a rhombic dodecahedron cut in half-   1350 rhombic dodecahedron puzzle-   1351 puzzle which is a rhombic dodecahedron cut in half-   1353 a puzzle except half a cube from a puzzle which is a rhombic    dodecahedron cut in half-   1400 skewed hexagonal prism-   1401 triangular prism-   1403 mirror image symmetrical triangular prism-   1420 skewed hexagonal prism-   1450 skewed hexagonal prism puzzle-   1500 elongated rhombic dodecahedron-   1501 helmet-shaped polyhedron-   1503 polyhedron cut away from helmet-shaped polyhedron 501-   1505 skewed triangular prism-   1507 mirror image of skewed triangular prism 505-   1550 elongated rhombic dodecahedron puzzle-   1551 helmet-shaped polyhedron puzzle-   1553 puzzle cut away from helmet-shaped polyhedron puzzle 551-   1600 truncated octahedron-   1601 enneahedron (diamond)-   1603 half of truncated octahedron-   1650 truncated octahedron puzzle-   1651 half of truncated octahedron puzzle-   1700 cube-   1701 hexahedron-   1703 polyhedron-   1705 cube with open hole-   1707 polyhedron formed by cutting through cube with open hole-   1709 polyhedron formed by cutting through half of cube-   1750 cube puzzle-   1751 puzzle formed by cutting through cube with open hole-   1753 half of cube puzzle

DETAILED DESCRIPTION OF THE INVENTION First Embodiment

When the interior of a regular polyhedron is filled perfectly by aplurality of convex polyhedrons, these convex polyhedrons can be calledthe component elements (here called “atoms”) of a regular polyhedron. Inother words, this means the regular polyhedron can be divided intoseveral atoms. Only a regular tetrahedron, a cube, a regular octahedron,a regular icosahedron and a regular dodecahedron exist as a regularpolyhedron.

However, there are limitless methods for dividing a regular polyhedroninto several convex polyhedrons. That is, there are limitless atoms forforming a regular polyhedron.

Thus, the inventors of the present invention keenly examined which atomsshould be adopted in order to be able to fill all the regularpolyhedrons with their atoms while reducing the number of differentatoms. As a result, a shape of the minimum number of atoms for fillingall the regular polyhedrons was discovered. The circumstances in whichthe inventors of the present invention discovered a shape of the minimumnumber of atoms for filling all the regular polyhedrons is explainedbelow.

First, a step which excludes a self-evident atom is explained. Forexample, a regular tetrahedron is examined as a polyhedron. Let thecenter of the regular tetrahedron be the peak, and four triangles on thesurface of the regular tetrahedron as the bottom surface and the regulartetrahedron is divided into four congruent triangular pyramids. Becausethe bottom surface of this triangular pyramid is a regular triangle,this triangular pyramid has threefold rotation symmetry and mirror imagesymmetry and these are further divided into six congruent triangularpyramids. Here, when the mirror image symmetrical convex parts aredefined as the same atom, it is possible to perfectly fill the interiorof regular tetrahedron using 24 atoms of single type.

It is possible to apply the same division method as the division methodof the regular tetrahedron stated above to a regular polyhedron. Thatis, a cube and regular octahedron are filled by 48 atoms and a regulardodecahedron and regular icosahedron are filled by 120 atoms. However,each of these atoms is obvious and effective in filling a specificregular polyhedron. However, when each of these atoms is diverted to anatom of another regular polyhedron, it instantly becomes an atom withbad efficiency.

The condition A below is attached in order to remove this obvious atom.

[Condition A]

It is possible to use any atom as at least two types of regularpolyhedron.

Then, when an efficient atom is adopted for a specific polyhedron, it isnecessary to create a method which can use this atom effectively in thefilling of another regular polyhedron. Here, it is preferred that asmany atoms as possible should be used to fill any one of the regularpolyhedrons. However, a supplement is also required. This is because acube is self-expanding, and when one filling method is discovered, it ispossible to easily increase the number of atoms by 2³, 3² etc. simply bylining it up down, left right. Consequently, a cube is examinedrestricted to a basic filling method which does not use the selfexpanding properties. In this way, filling all of the regularpolyhedrons using as many atoms of as few varieties as possible whilesatisfying [condition A] becomes a problem.

Furthermore, defining a mirror image symmetrical convex polyhedron asthe same variety of atom.

-   (1) because a regular polyhedron has various symmetries it is    predicted that a lot of mirror image symmetrical polyhedrons will be    used, (2) because it is common sense to identify a pair of reverse    diagrams in a tiling problem very similar to this filling problem

Five types of regular polyhedron are known, a regular tetrahedron, aregular hexahedron (cube), a regular octahedron, a regular dodecahedronand a regular icosahedron. In order to fill these regular polyhedronswith various atoms so as to satisfy [condition A], first, it isnecessary to investigate the mutual relationships between the five typesof regular polyhedron.

In the cube 200 as in FIG. 2( a), the upper face square is determined byABCD and the bottom face square by EFGH, the cube is cut by a planewhich passes through the three peaks A, C, F and a triangular pyramid201 which includes the peak B is cut away as in FIG. 2( b). Because thiscross section is a regular triangle, if a triangular pyramid whichincludes the peaks D, E, G is cut away by the same method, the remainingsolid ACFH becomes a regular tetrahedron 100 as shown in FIG. 2 (a). Thesix dash lines in FIG. 2 (a) show the regular tetrahedron 100 obtainedby this cutting away.

Next, as is shown in FIG. 2( b), four triangular pyramids 201 which arecut away form a bottom surface of a regular triangle and a side face ofa right isosceles triangle. The peaks B, C, D, E, G of these triangularpyramids 201 are gathered at one point, and when arranged so that thebottom face forms a square, a quadrangular prism 301 which has four sidefaces being all regular triangles is formed, shown in FIG. 3( a). Whenthe bottom face squares as shown in FIG. 3( b) of two of thesequadrangular prisms 301 are attached to each other (when a pair ofbottom face squares are glued together) a regular octahedron 300 isformed. One relationship between the cube 200, the regular tetrahedron100 and the regular octahedron 300 is obtained using this cuttingmethod.

Next, focusing on the triangular pyramid 201 shown in FIG. 2( b), as isshown in FIG. 4( a) three points Q, T, U are each placed at proportionson three edges AB, AC, AF which satisfy the following formula:

[formula 1]

AQ: QB=AT:TC=AU:UF=1:τ  (1)

Here, τ is a golden proportion, and as is common knowledge can beobtained by the formula (2):

[formula 2]

$\begin{matrix}{\tau = {\frac{\sqrt{5} + 1}{2} \approx 1.618033989}} & (2)\end{matrix}$

Then, as is shown in FIG. 4( b) a triangular pyramid 203 including thepeak A is cut away from the triangular pyramid 201. As is shown in FIG.5( a), (b), when the triangular pyramid 203 including the peaks C, F isalso cut away, a heptahedron 207 is left in the center. Here, the threepoints R, S, V are each placed on the three edges CB, FB, FC at aproportion which satisfies the formula (3) below:

[formula 3]

CR:RB=FS:SB=FV:VC=1:τ  (3

When a heptahedron 207 is created in the center of FIG. 5 b) using thismethod, the back face triangle TUV is a regular triangle and the threesaid face right triangles TQU, USV, VRT are congruent with the left handside right triangle when the back face regular triangle TUV is dividedin half left and right.

Next, the relationship established in the formula (4) below will befocused on.

[Formula 4]

QBS=RBQ=SBR=90°  (4)

Four pairs of this heptahedron 207 and its mirror image symmetricalheptahedron 209 are created. By attaching (gluing) these together, aregular icosahedron 400 is formed as shown in FIG. 5( c). As a result,one relationship is established between the triangular pyramid 201 andthe regular icosahedron 400.

Next, a regular dodecahedron 500 is shown in FIG. 6( a). As is shown inFIG. 6( b), the upper part is cut away by a plane which passes throughfour peaks A, B, C, D, and the right part is cut away by a plane whichpasses through four peaks A, B, E, F. Then, two cross sections becomecongruent squares and the two cut away solids become congruentpentahedrons 501. As is shown by the dash line in FIG. 6( c), whenfurther four planes are cut away, a cube 200 is left at the center. Onerelationship between the regular dodecahedron and the cube is obtained.

The regular tetrahedron 100, regular octahedron 300 and regularicosahedron 400 are obtained from dividing the cube 200 as a result ofthe above stated examination. In addition, the cube 200 is obtained fromdividing the regular dodecahedron 500. This dividing satisfies thecondition [condition A].

Next, it is examined how it is possible to fill the entire regularpolyhedron while satisfying condition [condition A] with as few types ofatoms as possible. The case is considered where the regular tetrahedron100 is formed from the dash line regular tetrahedron itself shown inFIG. 2( a). Next, the cube 200 is formed by three types of convexpolyhedron (five types if mirror images are included), namely oneregular tetrahedron 100, two heptahedrons 207 and two mirror imagesymmetrical heptahedrons 208, and six tetrahedrons 203 and six mirrorimage symmetrical tetrahedrons. Next, the regular octahedron 300 isformed by two types of convex polyhedron (four types if mirror imagesare included), namely four heptahedrons 207 and four mirror imagesymmetrical heptahedrons 208, and twelve tetrahedrons 203 and twelvemirror image symmetrical tetrahedrons. Next, the regular dodecahedron asshown in FIG. 6( b) is formed by four types of convex polyhedron (sixtype if mirror images are included), namely six pentahedrons 501, oneregular tetrahedron 100, two heptahedrons 207 and two mirror imagesymmetrical heptahedrons 208, six tetrahedrons 203 and six mirror imagesymmetrical tetrahedrons. Lastly, a regular icosahedron is formed by onetype of convex polyhedron consisting of four heptahedrons 207 and fourmirror image symmetrical heptahedrons 208.

From the examination above, it is clear that any of four types of convexpolyhedron (six types if mirror images are included) are used in forminga regular polyhedron, namely

[1] regular tetrahedron 100 (used for regular tetrahedron 100, cube 200and regular dodecahedron 500),

[2] heptahedron (equihepta) 207 (and its mirror image 208) (used in cube100, regular octahedron 300 and regular icosahedron 400),

[3] tetrahedron (golden tetra) 203 (and its mirror image 204) (used incube 100 and regular octahedron 300), and

[4] pentahedron (roof) 501 (used in regular dodecahedron 500).

If the four types of convex polyhedron are each divided into severalcongruent convex polyhedrons, the number of atoms which form the regularpolyhedron increases. Obvious dividing is removed and examined. First,in the regular tetrahedron 100 in [1], the regular tetrahedron 100 whichis extracted from FIG. 2( a) is shown again in FIG. 7( a) and theregular tetrahedron 100 is cut by a plane which passes through themidpoints I, J, K, L on the four edges AF, FC, CH, HA. Then, as is shownin FIG. 7( b), this cross section becomes a square (petri figure) andtwo solids which are divided become congruent prisms (triangular prisms)101. FIG. 7( c) shows a view from a different angle with one of theprisms 101 extracted. When the prism 101 is cut by a plane which passesthrough the three points J, L, M where M is at the center of the edge FHas shown in FIG. 7( d), two congruent pentahedrons 103 are formed.Moreover, because each pentahedron 103 is symmetrical, when the leftside pentahedron 103 is taken, it is possible to further divide thepentahedron into two congruent tetrahedrons 110 and 111 as is shown inFIG. 7( e). FIG. 8 shows a development of the right side tetrahedron 110(KJLH) and the dimensions are described as ratios. One atom which formsa regular polyhedron with the tetrahedron 110 as α is defined below.Then, as is shown in FIG. 1, because the regular tetrahedron 100 isformed by eight atoms, this can be written as α₈, as in a molecularformula.

Here, as is shown in FIG. 7( e) atom α is the tetrahedron (convexpolyhedron) 110 with peaks KJLH, and its mirror image tetrahedron 111with peaks MJLH. The atom α satisfies the conditions shown in formula(5) below.

[formula 5]

JL:LK:KL:LH:HL:LJ:JK:KH:JH=2:√2:√2:√2:√2:2:√2:√2:√6   (5)

Next, when the equihepta 207 in [2] is examined, the equihepta 207 shownagain in FIG. 9( a) has three rotational symmetry when the regulartriangle TUV is the bottom surface and the point B is the peak. As aresult, when the center of the regular triangle is O, and the equiheptais cut in a perpendicular direction by three faces divided into 120°pieces through O, the equihepta is always divided into three congruentsolids 210. This division is shown in FIG. 9( b) and a development isshown in FIG. 10.

One atom which forms a regular polyhedron with the heptahedron 210 as βis defined below. The atom β 210, as shown in FIG. 10, is a heptahedron(convex polyhedron) having edges BX, XY, YO, OA₁, A₁Z, ZT, TZ, ZA₁, A₁O,OY, YX, XR, RX, XB, BO, BZ, BR, RT, TY and TA₁.

Here, a, k, k′, b, c, d satisfy the following formulas (6) to (11)

(formula 6)

a=3−√5=2(2−τ)   (6)

(formula 7)

k=1/(τ+2)=(5−√5)/10   (7)

(formula 8)

k′=1−k=(5+√5)/10   (8)

(formula 9)

b=2/√3   (9)

(formula 10)

C=(2/τ²)√(τ²−2k′+3k′ ²)   (10)

(formula 11)

d=(2/τ²)√(4/3−4k′+4k′ ²)   (11)

At this time, atom β 210 satisfies the conditions shown in formula (12)below.

$\begin{matrix}\left( {{formula}\mspace{14mu} 12} \right) & \; \\{{{BX}:{{XY}\;: {{YO}: {{OA}_{1}: {A_{1} {Z:{{ZT}:{{TZ}:{{ZA}_{1}:{A_{1}{O:{{OY}:{{YX}:{{XR}:{{RX}:{{XB}:{{BO}:{{BZ}:{{BR}:{{RT}:{{TY}:{TA}_{1}}}}}}}}}}}}}}}}}}}}}} = {d:{{ak}:{c:{c:{{ak}:{\left. \sqrt{}3 \right. \cdot {{ak}:{\left. \sqrt{}3 \right. \cdot {{ak}:{{ak}:{c:{c:{{ak}:{{\left. \sqrt{}3 \right. \cdot {{ak}'}}:{{\left. \sqrt{}3 \right. \cdot {{ak}'}}:{d:{b:{d:{2 - {a:{a:{{2{{ak}'}}:{2{ak}}}}}}}}}}}}}}}}}}}}}}}}} & (12)\end{matrix}$

Because a regular icosahedron is formed by twenty-four atoms β 210, itis possible to describe this as β₂₄.

Next, when the golden tetra 203 in [3] is examined, it is impossible todivide the golden tetra 203 into a plurality of congruent convexpolyhedrons. Thus, one atom which forms a regular polyhedron with thegolden tetra 203 as γ is defined. FIG. 11 is a development of the atom γ203 and dimensions are shown according to ratios. The atom γ, as isshown in FIG. 11, is a tetrahedron (convex polyhedron) having edges CR,RC, CV, VC, CT, TC, RT, RV and VT. The atom γ satisfies the formula (13)below.

(formula 13)

CR:RC:CV:VC:CT:TC:RT:RV:VT =a:a:√3a:√3a:a:a:√2a:2a: √2(√5−1)   (13)

Then, because a regular octahedron is formed by twenty-four β and γ, itis possible to describe this as β₂₄γ₂₄. In addition, because a cube inFIG. 1( a) is formed by eight α, twelve β and twelve γ, it is possibleto describe this as α₈β₁₂γ₁₂.

Next, when the roof 501 in [4] is examined, while considering that theroof 501 has up down left and right symmetry, it is divided into twocongruent pentahedrons 501 as is shown in FIG. 12( a). One atom whichforms a regular polyhedron with the pentahedron 510 as δ is defined.FIG. 13 is a development of the atom δ 510 and dimensions are shownaccording to ratios. The atom δ 510, as is shown in FIG. 13, is atetrahedron (convex polyhedron) having edges BE, ED₁, D₁A, AD₁, D₁E, EB,BC₁, C₁E, C₁D₁, BA and AE. The atom δ 510 satisfies the formula (14)below.

(formula 14)

BE:ED ₁ :D ₁ A:AD ₁ :D ₁ E:EB:BC ₁ :C ₁B: C₁ E:C ₁D:BA:AE=2:√(4−τ):√(4−τ):√(4−τ): √(4−τ):2:2(τ−1):2(τ−1):2(τ−1): τ−1:2:2√2  (14)

A solution for forming a regular polyhedron by a minimum number of atomsα, β, γ and δ is obtained as described above. The number of each regularpolyhedron using the atoms α, β, γ, δ can be expressed as:

Regular tetrahedron α₈ Cube α₈β₁₂γ₁₂ Regular octahedron β₂₄γ₂₄ Regulardodecahedron α₈β₁₂γ₁₂δ₁₂ Regular icosahedron β₂₄

However, a mirror image symmetrical atom is also shown by α, β, γ, δ. Ifa prime symbol (′) is attached to distinguish a mirror image symmetricalatom, then they can be expressed as:

Regular tetrahedron α₄α′₄ Cube α₄α′₄β₆β′₆γ₆γ′₆ Regular octahedronβ₁₂β′₁₂γ₁₂γ′₁₂ Regular dodecahedron α₄α′₄β₆β′₆γ₆τ′_(γ)δ₁₂ Regularicosahedrons β₁₂β′₁₂

From the above, it was discovered that the minimum number of atoms forforming a regular polyhedron is four types α, β, γ, δ.

Here, it is clear that the minimum number of atoms for forming a regularpolyhedron can be expressed as theorem 1 under the definitions (1)-(4)below.

-   (1) polyhedrons P and Q are “congruent” means either that P and Q    are identical or that P and Q have a mirror image relationship.-   (2) a polyhedron is “indecomposable” means that P is indivisible    into two or more congruent polyhedrons.-   (3) Let II be a group of polyhedrons P₁, P₂, . . . P_(n).-   i.e. II={P₁, P₂, . . . P_(n)}-   Let E be a group of indecomposable polyhedrons e₁, e₂, . . . e_(m),-   i.e. E={c₁, c₂, . . . c_(m)}, ∀e₁ is indecomposable, c_(i), and    c_(j) are non-congruent (i≠j)-   At this time, E is a group E (II) of the element (atom) of II means    satisfying formula (15) below.

(formula 15)

(∀_(i)<1≦i≦n), a_(i) is a nonnegative integer.   (15)

That is, whichever polyhedron P_(i) belongs to II, it is possible todivide into an indecomposable polyhedron belonging to E.

-   (4) The element number (atom number) e (II) of II means the minimum    number of an order among various element groups with respect to II.-   That is e (II)=min|E (II)|

(Theorem 1)

-   Let II₁={regular polyhedron group}-   e (II₁)≦4, e (II₁)={α, β, γ, δ}

Here, a three-dimensional puzzle of the present invention related to thepresent embodiment which uses the minimum number of atoms α, β, γ, δ forforming all the regular polyhedrons will be explained in detail. Thethree-dimensional puzzle of the present invention related to the presentembodiment creates a regular polyhedron using the minimum number ofatoms α, β, γ, δ, and has the excellent effects of being able tovisually explain the above stated theorem 1. Consequently, thethree-dimensional puzzle of the present invention related to the presentembodiment is excellent as an educational material for explaining theabove stated theorem 1.

(Regular Tetrahedron Three-Dimensional Puzzle)

FIG. 1 is referenced. FIG. 1 is a three-dimensional puzzle of thepresent invention related to the present embodiment and shows a regulartetrahedron three-dimensional puzzle 100. As is shown in FIG. 1, thethree-dimensional puzzle (regular tetrahedron three-dimensional puzzle100) can be formed by four pieces which have the shape of atom α 110 andfour pieces which have a shape of a mirror image α′ 111 of atom α 110 asexplained in detail above. Furthermore, the regular tetrahedronthree-dimensional puzzle 100 of the present invention related to thepresent embodiment is one example and there are also examples where thearrangement of the atom α 110 and the atom α′ 111 can be changed whilemaintaining the relative positional relationship between them.

(Cube Three-Dimensional Puzzle)

FIG. 14 is referenced. FIG. 14 is a three-dimensional puzzle of thepresent invention related to the present embodiment and shows a cubethree-dimensional puzzle 200. The cube includes the regular tetrahedron100 which is comprised of the atom α (and α′) as explained in detailabove, and is formed by gluing a triangular pyramid 201 which has sidefaces which are right isosceles triangles to each of the four faces ofthe cube. In addition, as is shown in FIG. 15, the triangular pyramid201 is a part arranged with three pieces 203 which has the shape of theatom γ with parts 207 (FIG. 16) comprised of three pieces 210 which havethe shape of atom β at the center. Therefore, the three-dimensionalpuzzle of the present invention related to the present embodiment (cubepuzzle 200) can be formed by using four pieces 110 which have the shapeof atom α and four pieces 111 which have the shape of the atom α mirrorimage atom α′, six pieces 210 which have the shape of atom β and sixpieces 211 which have the shape of the atoms β mirror image atom β′, andsix pieces 203 which have the shape of atom γ and six pieces 204 whichhave the shape of the atom γ mirror image atom γ′. Furthermore, the cubethree-dimensional puzzle 200 of the present invention related to thepresent embodiment is one example and there are also examples where thearrangement of the atoms α and α′, β and β′, and γ and γ′ can be changedwhile maintaining the relative positional relationship between them.

(Regular Octahedron Three-Dimensional Puzzle)

FIG. 17 is referenced. FIG. 14 is a three-dimensional puzzle of thepresent invention related to the present embodiment and shows a regularoctahedron three-dimensional puzzle 300. The regular octahedronthree-dimensional puzzle 300 is formed by gluing eight regular threesided pyramids 201 which have sides which are right isosceles triangles.Therefore, the three-dimensional puzzle of the present invention relatedto the present embodiment (regular octahedron puzzle 300) can be formedby using twelve pieces 210 which have the shape of atom β and twelvepieces 211 which have the shape of the atom β mirror image atom β′, andtwelve pieces 203 which have the shape of atom γ and twelve pieces 204which have the shape of the atom γ mirror image atom γ′. Furthermore,the regular octahedron three-dimensional puzzle 300 of the presentinvention related to the present embodiment is one example and there arealso examples where the arrangement of the atoms β and β′, and γ and γ′can be changed while maintaining the relative positional relationshipbetween them.

(Regular Dodecahedron Three-Dimensional Puzzle)

FIG. 18 is referenced. FIG. 18 is a three-dimensional puzzle of thepresent invention related to the present embodiment and shows a regulardodecahedron three-dimensional puzzle 500. The regular dodecahedronthree-dimensional puzzle 500 includes the cube 200 at the center asshown in FIG. 6( c), and is formed by arranging roof parts 501 comprisedof two pieces 510 which have a shape of the atom δ on six sides of cubes200. Therefore, the three-dimensional puzzle of the present inventionrelated to the present embodiment (regular dodecahedron puzzle 500) canbe formed by using four pieces 110 which have the shape of atom α andfour pieces 111 which have the shape of the atom α mirror image atom α′,six pieces 210 which have the shape of atom β and six pieces 211 whichhave the shape of the atoms β mirror image atom β′, six pieces 203 whichhave the shape of atom γ and six pieces 204 which have the shape of theatom γ mirror image atom γ′, and twelve pieces 510 which have the shapeof the atom δ. Furthermore, the regular dodecahedron three-dimensionalpuzzle 500 of the present invention related to the present embodimentshown in FIG. 18 is one example and there are also examples where thearrangement of the atoms α 110 and α′111, β 210 and β 211′, γ 203 and γ′203, can be changed while maintaining the relative positionalrelationship between them.

(Regular Icosahedron Three-Dimensional Puzzle)

FIG. 19 is referenced. FIG. 19 is a three-dimensional puzzle of thepresent invention related to the present embodiment and shows a regularicosahedron three-dimensional puzzle 400. Four pairs of parts 207 whichhave the shape of an equihepta comprised from three pieces 210 whichhave the shape of the atom β shown in FIG. 16, and the parts 209 whichhave the equihepta mirror image symmetrical shape are glued together toform the regular icosahedron. Therefore, the three-dimensional puzzle ofthe present invention related to the present embodiment (regularicosahedron puzzle 400) can be formed by using twelve pieces 210 whichhave the shape of atom β and twelve pieces 211 which have the shape ofthe atoms β mirror image atom β′. Furthermore, the regular icosahedronthree-dimensional puzzle 400 of the present invention related to thepresent embodiment shown in FIG. 19 is one example and there are alsoexamples where the arrangement of the atoms β 210 and β′ 211, can bechanged while maintaining the relative positional relationship betweenthem.

As explained above, because the three-dimensional puzzle of the presentinvention related to the present embodiment can be formed by a minimumnumber of atoms (convex polyhedrons) which fill all the regularpolyhedrons, the present invention demonstrated excellent effects ofbeing able to form all the regular polyhedrons (regular tetrahedron,cube, regular octahedron, regular dodecahedron, regular icosahedron)even with few parts. In addition, it is possible to visually prove thetheorem discovered by the inventors and the present invention can alsobe used as an excellent educational material for explaining the abovestated theorem 1.

Second Embodiment

In the second embodiment, an example of a three-dimensional puzzle whichcan form a cube and a regular dodecahedron without using the atom α ofembodiment one is explained.

FIG. 20 is referenced. FIG. 20( a) shows the appearance when six roofs501 (δ²) comprised from two atoms δ 510 are reversed so as to fill theinterior of a cube 200. In FIG. 20( a) only the roof 501 of the backface, side face and bottom face is shown to facilitate visualization.Each face of the pentahedron 501 (δ₂) which is reversed overlaps so thatno gaps exist as is clear from the dihedral angle of each face. However,a long thin gap is produced in the eight corners of the cube and fourmirror image symmetrical hexahedrons 601 and four hexahedrons 602 arerequired to fill these gaps. FIG. 20( b) shows a hexahedron 601 forfilling the forefront left side lower gap in FIG. 20( a), and FIG. 20(c) shows a development of this. In addition, FIG. 20( d) shows theappearance when two lower left gaps of two of the mirror imagesymmetrical hexahedrons 601 and 602 are filled. In this way, arelationship between a regular dodecahedron 500 and cube 200 is obtainedwhich is different to the relationship in the first embodiment.

A cube which is different to the cube explained in the first embodimentis comprised of two types of convex polyhedron, namely, six roofs 501(δ₂), four hexahedrons 601 and four mirror image symmetrical hexahedrons602. In addition, the regular dodecahedron 500 is comprised of two typesof convex polyhedron, namely, twelve roofs 501 (δ₂), four hexahedrons601 and four mirror image symmetrical hexahedrons 602.

From the above investigation it is clear that the five types of convexpolyhedrons are used in the structure of any two or more regularpolyhedrons:

-   [1] regular tetrahedron 100 (used in a regular tetrahedron 100 and    cube 200).-   [2] equihepta 207 and its mirror image 209 (used in a cube 100,    regular octahedron 300 and regular icosahedron 500).-   [3] golden tetra 203 and its mirror image 204 (used in a cube 200    and regular octahedron 300).-   [4] roof 501 (used in a regular dodecahedron 500).-   [5] hexahedron 601 and its mirror image 602 (used in a cube 200 and    regular dodecahedron 500).

Here, because the convex polyhedrons [1] to [4] were explained in thefirst embodiment, an explanation is omitted here.

When the hexahedron 601 in [5] is examined, it has three rotationalsymmetry as is clear from the hexahedron in FIG. 20( b) and is thedevelopment in FIG. 20( c). As a result, by cutting in a perpendiculardirection by three faces of 120° each, the hexahedron is divided intothree congruent polyhedrons. A development of three congruentquadrangular pyramids 610 which have been cut along the edges of thehexahedron 601 is shown in FIG. 21. Furthermore, dimensions aredescribed using ratios in the development. A quadrangular pyramid 610shown in FIG. 21 is defined as atom ε.

Atom ε 610 is a pentahedron (convex polyhedron) which has edges D₁E₁,E₁F₁, F₁G₁, G₁H₁, H₁I₁, I₁J₁, J₁K₁, K₁D₁, D₁F₁, D₁I₁, D₁J₁ and F₁I₁ Atomε satisfies the following formula (16)

(formula sixteen)

D ₁ E ₁ :E _(i) F ₁ :F ₁ G ₁ :G ₁ H ₁ :H ₁ I ₁ :I ₁ J ₁ : J ₁ K ₁ :K ₁ D₁ :D ₁ F ₁ :D ₁ I ₁ :D ₁ J ₁ :F ₁ I ₁=√3:2−τ:2−τ:2−τ:√(18−11τ):√(18−11τ):2−τ:√3:√(4−τ):2(τ−1): √(4−τ):τ−1   (16)

Because the cube 200 is comprised of twelve atoms δ 510 and twenty-fouratoms ε (twelve atoms ε and twelve mirror image atoms ε′), it isexpressed as δ₁₂ε₂₄ (δ₁₂ε₁₂ε′₁₂) and because the regular dodecahedron500 is comprised of twenty-four δ and twenty-four ε, it is expressed asδ₂₄ε₂₄ (δ₂₄ε₁₂ε′₁₂).

By using the atoms α, β, γ, δ, ε, each individual regular polyhedronbecomes:

Regular tetrahedron α₈ Cube α₈β₁₂γ₁₂, δ₁₂ε₂₄ Regular octahedron β₂₄γ₂₄Regular dodecahedron δ₂₄ε₂₄However, the same symbols α, β, γ, δ, ε, also indicate mirror imagesymmetrical atoms. If the mirror image symmetrical atoms are attachedwith a prime symbol (′) then:

Regular tetrahedron α₄α′₄ Cube α₄α′₄β₆β′₆γ₆γ′₆, δ₁₂ε₁₂ε′₁₂ Regularoctahedron β₁₂β′₁₂γ₁₂γ′₁₂ Regular dodecahedron δ₂₄ε₁₂ε′₁₂ Regularicosahedron β₁₂β′₁₂

In the three-dimensional puzzle of the present invention related to thepresent embodiment, it is possible to form a cube and a regulardodecahedron without using atom α by using ε in addition to the minimumnumber of atoms α, β, γ, δ which form all the regular polyhedrons.Consequently, the three-dimensional puzzle of the present inventionrelated to the present embodiment is created from regular polyhedronsusing the atom ε in addition to the minimum number of atoms α, β, γ, δ,it is possible to visually explain the above stated theorem 1 anddemonstrate excellent effects which can compare both by a transformationexample using ε which is the fifth atom. Therefore, thethree-dimensional of the present invention related to the presentembodiment is also excellent as an educational material.

Third Embodiment

In the present embodiment, an example of a three-dimensional puzzlewhich can form a regular tetrahedron 100, cube 200 and regularoctahedron 300 without using the atom α of the first embodiment isexplained.

FIG. 22 is referenced. FIG. 22( a) shows the atom α 110 shown in FIG. 7(e) seen from a different direction. Next, the atom α 110 is divided intwo as in FIG. 22( b) from the peak L perpendicular to the edge JH withthe foot being W. Then, the left side tetrahedron 121 becomes threeportions by dividing the regular tetrahedron 100 in a perpendiculardirection from the peak and this can be divided into two mirror imagesymmetrical tetrahedrons 151 and 152 as shown in FIG. 22( c). Thetetrahedron 151 shown in FIG. 22( c) is an atom and defined as ξ. Adevelopment of the atom ξ 151 is shown in FIG. 23. Each dimension isshown by ratios. As is shown in FIG. 22( c), the atom ξ 151 is atetrahedron (convex polyhedron) which has edges LM₁, M₁L₁, L₁M₁, M₁J,JM₁, M₁L, LL₁, LJ and L₁J. The atom ξ satisfies the following formula(17). Furthermore, as is shown in FIG. 22( c) the atom ξ has a mirrorsymmetrical atom ξ′ 152.

(formula 17)

LK ₁ :M ₁ L ₁ :L ₁ M ₁ :M ₁ J:JM ₁ :M ₁ L:LL ₁ ,:LJ:L ₁ J=2/√3:1/√6:1/√6:√(2/3):√(2/3):√(2/3):2√3: √(3/2):√2:1/√2   (17)

In addition, with respect to the right side tetrahedron 123 in FIG. 22(b), a mirror image symmetrical tetrahedron 124 is created and iscombined (glued) by the face of triangle HLW. FIG. 22( d) is aquadrangular pyramid obtained from this method and is attached togetherby an unforeseen shape with the atom β 210 and atom γ 203.

In order to show this, first the atom γ 203 is focused on. This is aslightly long and triangular pyramid, however, as is shown in FIG. 24,the atom θ 800 is cut from the end of the atom γ. Then, a development ofeach of the polyhedrons is shown in FIG. 25 and FIG. 26. The polyhedronin which the atom θ 800 is cut from the end of the atom γ is defined asatom η 900. In each of FIG. 25 and FIG. 26 which are developments, thedimension of each edge is shown as a ratio. Furthermore, the atom η 900and the atom θ 800 each has a mirror symmetrical atom η′ 901 and θ′ 801.

Atom θ 800 is a tetrahedron (convex polyhedron) which has edges VO₁,O₁P₁, P₁O₁, O₁N₁, N₁O₁, O₁V, VP₁, VN₁ and N₁P₁ as is shown in FIG. 25.Furthermore, the atom θ has a mirror symmetrical atom θ′ 801. The atom θ800 satisfies the formulas (18) and (19) below.

(formula 18)

e=√(2/5(9−4√5))   (18)

(formula 19)

VO ₁ :O ₁ P ₁ :P ₁ O ₁:O₁ N ₁ :N ₁O₁ :O ₁ V: VP ₁ :VN ₁ :N ₁ P₁=√3ak:e:e:ak:ak:√3ak√2(√5−2):2ak:√3e   (19)

Atom η 900 is a pentahedron (convex polyhedron) which has edges O₁N₁,N₁O₁, O₁P₁, P₁O₁, O₁C, CT, TC, CR, RC, CO₁, N₁P₁, P₁T, TR and N₁R, as isshown in FIG. 25. Furthermore, the atom η 900 has a mirror symmetricalatom η′ 901.

FIG. 26 is a development of the atom η 900 and satisfies the formula(20) below.

(formula 20)

O ₁ N ₁ :N ₁ O ₁ :O ₁ P ₁ :P ₁ O ₁ :O ₁ C:CT:TC:CR:RC: CO ₁ :N ₁ P ₁ :P₁ T:TR:N ₁ R=ak:ak:e:e;√3ak′:a:a:a:a:√3ak′: √3e:√2:2a:2ak′  (20)

By combining one of three types of atom, β 210, η 900 and θ 800, it ispossible to make a quadrangular pyramid 701 shown in FIG. 22( d).

By adding four atoms ξ 151 to the quadrangular pyramid 701, two atoms α110, that is, the polyhedron 700 shown in FIG. 27 is obtained. As statedabove, the regular tetrahedron is formed by sixteen ξ 151, four β 210,four η 900 and four θ 800, in total β₄ξ₁₆η₄θ₄ (β₂β′₂ξ₈ξ′₈η₂η′₂θ₂θ′₂).Furthermore, as already indicated, the regular tetrahedron 200 is alsoformed by ξ₆.

When a regular tetrahedron is formed by this method, the entirestructure using the atom α 110 is replaced with this, and the regulartetrahedron 100, the cube 200, the regular octahedron 300 which areformed by the atoms β 210, ξ 151, η 900 and θ 801 each become:

Regular tetrahedron β₄ξ₁₆η₄θ₄ξ₆ (β₂β′₂ξ₈ξ′₈η₂η′₂θ₂θ′₂ξ₃ξ′₃) Cubeβ₁₆ξ₁₆η₁₆θ₁₆ (β₈β′₈ξ₈ξ′₈η₈η′₈θ₈θ′₈) Regular octahedron β₂₄η₂₄θ₂₄(β₁₂β′₁₂η₁₂η′₁₂θ₁₂θ′₁₂)In this way, the number of atoms which are used in the structureincreases significantly, however, the type of atom only increases byone.

In the three-dimensional puzzle of the present invention related toembodiment 1, an example in which it is possible to make a structurewith a minimum number of atoms α, β, γ, δ which form all the regularpolyhedrons is explained. However, the three-dimensional puzzle of thepresent invention related to the present embodiment can form a regulartetrahedron, cube and regular dodecahedron without using the atom α.Consequently, because the three-dimensional puzzle of the presentinvention related to the present embodiment can form the atom α bycombining different atoms and form a regular tetrahedron, cube andregular dodecahedron without using the atom α, it is possible tovisually explain the above stated theorem 1 and demonstrate excellenteffects which can compare both by a transformation example using ξ, θ, ηwhich are the sixth to eighth atoms. Therefore, the three-dimensionalpuzzle of the present invention related to the present embodiment isalso excellent as an educational material.

Fourth Embodiment

In the fourth embodiment, an example is explained in which thethree-dimensional puzzle which has the atoms explained in the first tothird embodiments as structural components is applied to a Fedorovpolyhedron or a Fedorov space filling solid. As stated previously, in aconvex polyhedron, convex polyhedrons P and Q are congruent if eitherthe two convex polyhedrons P and Q are the same or when they have arelationship where one of the convex polyhedrons is a mirror image ofthe other.

When a non-congruent convex polyhedron P_(i) is 1≦i≦n, then II={P₁, P₂,. . . , P_(n)}. If at least one element P which is included in P_(i), isformed by face to face joining of congruent convex polyhedrons of convexpolyhedron σ, then convex polyhedron σ is called an atom of II.

In addition, a parellelohedron is a convex polyhedron which tilesthree-dimensional space using a face to face joining of its translates.Minkowski obtained the following results for a general d-dimensionalparallelohedron.

(Minkowski Theorem A)

-   If P is a d-dimensional parallelohedron, then-   1) P is centrally symmetric,-   2) All faces of P are centrally symmetric,-   3) The projection of P along any of its (d-2) faces onto the    complementary 2-plane is either a parallelogram or a centrally    symmetric hexagon.

(Minkowski Theorem B)

The number f_(d-1) (P) of faces of a d-parallelohedron P does not exceed2(2^(d)−1) and there is a parallelohedron P with f_(d-1)=2(2^(d)−1).

Dolbilin extended Minkowski's theorems for non-face to face tilings ofspace. There are also numerous studies on parallelohedra discussed byAlexandrov and Gruber.

In 1890, a Russian crystallographer, Evgraf Fedorov, established thatthere are exactly five types of parallelohedra, namely, parallelopiped(cube), rhombic dodecahedron, skewed hexagonal prism (parallel hexagonalprism), elongated rhombic dodecahedron and truncated octahedron shown inFIG. 29( a) to FIG. 29( e). These parallelohedra are called “Fedrovpolyhedrons” or “Fedrov space filling solids”. FIG. 30( a) to FIG. 30(e) shows the space filling characteristics of each type ofparallelohedron.

The characteristics which are common to these space filling solids isnot only that any one type of solid can fill space without any gaps, butalso that any two solids can be overlapped by only translation of it asfilling the space of each solid.

Let II₁ be a group consisting of a cube, skewed hexagonal prism, rhombicdodecahedron, elongated rhombic dodecahedron and truncated octahedron.The inventors keenly examined which atom should be adopted for fillingall the space filling solids included in II₁ by the minimum number ofatoms. As a result, the shape of the minimum number of atoms for fillingall the space filling solids was discovered. Next, the circumstances inwhich the inventors discovered the shape of the minimum number of atomsfor filling all the space filling solids is explained. In addition, inthe present embodiment, a three-dimensional puzzle having convexpolyhedrons which form five types of Fedrov space filling solids isexplained.

(Cube and Rhombic Dodecahedron)

The cube 1200 in FIG. 31( a) is decomposed into six congruentquadrangular pyramids 1201 shown in FIG. 31( b) using six planes, eachof which passes through the center of the cube and contains two oppositeedges. The quadrangular pyramid 1201, as shown in FIG. 31( a), has asquare base and the four sides are congruent isosceles triangles.

The quadrangular pyramid 1201 is further divided into four congruenttriangular pyramids 1203 as shown in FIG. 32( b) by two perpendicularplanes passing through the peak and through each of the two diagonals ofthe base. When these triangular pyramids 1203 are shown again in FIG.32( c), because three faces are orthogonal at one peak, the triangularpyramid 1203 is called a right tetra 1203. Thus, the cube 1200 can bedivided into twenty-four congruent right tetra 1203. If the process isreversed, then it is possible to form the cube 1200 by twenty-fourcongruent right tetra 1203.

In addition, as is shown in FIG. 33( a) and FIG. 33( b), by attaching aquadrangular pyramid 1201 to each of the six congruent faces of the cube1200, a rhombic dodecahedron 1300 is formed. This shows that it ispossible to form a rhombic dodecahedron 1300 from forty-eight congruentright tetra 1203. In this way, the right tetra 1203 is a commonconstituent element of the cube 1200 and the rhombic dodecahedron 1300.

(Skewed Hexagonal Prism)

As is shown in FIG. 32( c), the base of the right tetra 1203 is a rightisosceles triangle. Two right tetras 1203 which are glued at the base ofthese right isosceles triangles form another tetrahedron 1100 as isshown in FIG. 34( a). This tetrahedron is called a Sphenoid 1100. InFIG. 34( a) the dash line represents the gluing surface of twotriangular pyramids.

It is known that the sphenoid 1100 has noteworthy characteristics whichare called self-expanding. This is because when eight sphenoid 1100 arecombined, the length of each edge expands to exactly twice that of asimilar triangular pyramid which makes it easy to understand that thesphenoid is a space filling solid.

Moreover, when three sphenoid 1100 are combined it is possible to form atriangular prism 1401 with a regular triangle as a perpendicular crosssection shown in FIG. 34( b). Then, the skewed hexagonal prism 1400 ofFedorov shown in FIG. 34( c) is formed using three pairs of thetriangular prism 1401 and its mirror image symmetrical triangular prism1403 so that the upper and lower faces are adjusted to form parallelhexagonals. Therefore, a skewed hexagonal prism 1400 can be formed byeighteen sphenoids 1100.

(Elongated Rhombic Dodecahedron)

Next, FIG. 35( a) shows an elongated rhombic dodecahedron 1500. Thiselongated rhombic dodecahedron 1500 is formed from two parts. One partis a rhombic dodecahedron 1300 and the other part is a concavepolyhedron 1501 which can be obtained from the rhombic dodecahedron. Inorder to obtain the second convex polyhedron 1501, we consider fourpeaks of the rhombic dodecahedron 1300. The elongated rhombicdodecahedron 1500 is cut along the edges containing these peaks shown inFIG. 35, and the elongated rhombic dodecahedron 1500 is opened at thispeak. The obtained polyhedron 1501 is formed in the shape of a helmet.As stated previously, the rhombic dodecahedron 1300 consists offorty-eight right tetras 1203. Therefore, the elongated rhombicdodecahedron 1500 can be formed using ninety-six right tetras 1203.

(Truncated Octahedron)

The sphenoid 1100 has another important property. As is shown in FIG.36( a), it can be divided into four congruent hexahedrons 1105 as isshown in FIG. 36( b) by using six planes wherein each plane passesthrough the midpoint of an edge of the sphenoid 1100 and must also beperpendicular to the edge. Each of these hexahedrons 1105 is called ac-squadron.

By arranging these four c-squadrons in an appropriate way, it ispossible to form an enneahedron (diamond) 1601 as shown in FIG. 37( a).Then, while adjusting so that a square and a regular hexahedron appearon the surfaces, a truncated octahedron 1600 is formed as shown in FIG.37( b) by combining (glued) together six diamonds 1601. In other words,the c-squadron 1105 is a constituent element of the truncated octahedron1600 and the truncated octahedron 1600 can be formed using twenty-fourc-squadrons 1105.

From the above, it can be seen that the cube (parallel hexagram) 1200can be formed by twenty-four right tetras 1203, the rhombic dodecahedron1300 can be formed by forty-eight right tetras 1203, the skewedhexagonal prism 1400 can be formed by eighteen sphenoids 1100, theelongated rhombic dodecahedron 1500 can be formed by ninety-six tetrasand the truncated octahedron can be formed by twenty-four c-squadrons1105.

Next, an examination is made whether common constituent elements existbetween the right tetra 1203, sphenoid 1100 and c-squadron 1105.

Here, one among the four c-squadrons 1105 shown in FIG. 36( b) is shownagain in FIG. 38( a). The c-squadron 1105 is a solid with left and rightsymmetry and is divided into two congruent pentahedrons 1101 and 1103 asis shown in FIG. 38( b). Then, as is shown in FIG. 38( c), by using twoeach of these new pentahedrons 1101 and 1103, it is possible to form theright tetra 1203 shown in FIG. 32( c).

In addition, as is shown in FIG. 1, it is possible to form the sphenoid1100 with eight of these pentahedrons (four pentahedrons 1101 and fourpentahedrons 1103). That is, the pentahedrons 1101 and 1103 are a commonconstituent element of the five types of Fedrov space filling solid.

It is not possible to divide the pentahedrons 1101 and 1103 into furthercongruent solids. The pentahedrons 1101 and 1103 are atoms of the fivetypes of Fedrov space filling solids, and the pentahedron 1101 isdefined as a and its mirror image 1103 is defined as σ′. A developmentincluding dimensions of the atom σ 1101 and atom σ′ 1103 is shown inFIG. 39. As is shown in FIG. 39, the atom σ is a pentahedron (convexpolyhedron) which has edges ab, bc, ca, cf, fc, ca, ad, da, ab, be, eb,de, ef, and fd, and the atom σ′ is a pentahedron (convex polyhedron)which has edges a′b′, b′c′, c′a′, c′f′, f′ c′ c′a′, a′d′, d′ a′ a′b′,b′e′, e′ b′ d′e′, e′ f′ and f′d′. The atom σ satisfies the followingformula (21). Furthermore, the formula of the atom σ′ which is themirror image of the atom σ is omitted.

(formula 21)

ab, bc, ca, cf, fc, ca, ad, da, ab, be, eb, de, ef,fd=2:√2:√2:√6:√6:√2:√2:√2:2:4:4:3√2: 2√3:√6   (21)

By using the atom σ 1101 and atom σ′ 1103, it is clear that the cube(parallel hexahedron) 1200 can be formed by ninety-six atoms σ(forty-eight atoms σ 1101 and forty-eight atoms σ′ 1103), the rhombicdodecahedron 1300 can be formed by one hundred and ninety-two atoms σ(ninety-six atoms σ 1101 and ninety-six atoms σ′ 1103), the truncatedoctahedron 1600 can be formed by forty-eight atoms σ (twenty-four atomsσ 1101 and twenty-four atoms σ′ 1103), the skewed hexagonal prism 1400can be formed by one hundred and forty-four atoms σ (seventy-two atoms σ1101 and seventy-two atoms σ′ 1103) and the elongated rhombicdodecahedron 1500 can be formed by three-hundred and eighty-four atoms σ(one-hundred and ninety-two atoms σ 1101 one-hundred and ninety-twoatoms σ′ 1103).

In addition, a cube is formed by ninety-six atoms σ 1101 and atoms σ′1103, however, it is possible to form a cube with as little astwenty-four atoms σ (twelve atoms δ 1101 and twelve atoms σ′ 1103).

From the above, it was discovered that the minimum number of atoms forforming a Fedrov space filling solid is one type of atom σ (and itsmirror image σ′).

Here, it is clear that it is possible to express the minimum number ofatoms which form the Fedrov space filling solids as theorem 2 under thedefinitions (1) to (4) below.

-   (1) The polyhedrons P and Q are “congruent” means that P and Q are    either exactly the same shape or are in a mirror image relationship.-   (2) A polyhedron is “indecomposable” means P are indivisible into    two or more congruent polyhedrons.-   (3) Let II be a group of polyhedrons P₁, P₂, . . . P_(n).-   i.e. II={P₁, P₂, . . . P_(n)}-   Let E be a group of indecomposable polyhedrons e₁, e₂, . . . e_(m),-   i.e. E={e₁, e₂, . . . e_(m)}, ∀e₁ is indecomposable, e_(i) and e_(j)    are non-congruent (i≠j)-   At this time, E is a group E (II) of the element (atom) of II means    satisfying formula (15) below.

$\begin{matrix}\left( {{formula}\mspace{14mu} 22} \right) & \; \\{{{P\; i} = {\sum\limits_{i = 1}^{m}\; {a_{i}e_{i}\; \left( {\forall_{i}{< 1 \leq i \leq n}} \right)}}},{a_{i}\mspace{14mu} {is}\mspace{14mu} a\mspace{14mu} {nonnegative}\mspace{14mu} {{integer}.}}} & (22)\end{matrix}$

That is, whichever polyhedron P_(i) belongs to II, it is possible todivide into an indecomposable polyhedron belonging to E.

-   (4) The element number (atom number) e (II) of II means the minimum    number of an order among various element groups with respect to II.-   That is e (II)=min|E (II)|

(Theorem 2)

-   Let II₂={a group of Fedrov space filling solids}-   e (II₂)≦1, E (II₂)={σ}

Here, the three-dimensional puzzle of the present invention related tothe present embodiment which uses a minimum number of atoms σ to formall the Fedrov space filling solids will be explained in detail. Thethree-dimensional puzzle of the present invention related to the presentembodiment creates Fedrov space filling solids using a minimum number ofatoms σ and demonstrates excellent effects of being able to visuallyexplain the above stated theorem 2. Consequently, the three-dimensionalpuzzle of the present invention related to the present embodiment isexcellent as an educational material for explaining the above statedtheorem 2.

(Regular Tetrahedron Three-Dimensional Puzzle)

FIG. 28 is referenced. FIG. 28 is a three-dimensional puzzle of thepresent invention related to the present embodiment and shows a sphenoidthree-dimensional puzzle. As is shown in FIG. 28, the three-dimensionalpuzzle (sphenoid) of the present invention related to the presentembodiment can be formed using four pieces which have a shape of an atomσ 1101 and four pieces which have a shape of the atom σ 1101 mirrorimage σ′ 1103 as explained above in detail.

(Parallel Hexagram Three-Dimensional Puzzle)

The parallel hexagram three-dimensional puzzle of the present inventionrelated to the present embodiment can be formed using forty-eight pieceswhich have a shape of an atom σ 1101 and forty-eight pieces which have ashape of the atom σ 1101 mirror image σ′ 1103 as explained above indetail.

(Skewed Hexagonal Prism Three-Dimensional Puzzle)

The skewed hexagonal three-dimensional puzzle of the present inventionrelated to the present embodiment can be formed using seventy-two pieceswhich have a shape of an atom σ 1101 and seventy-two which have a shapeof the atom σ 1101 mirror image σ′ 1103 as explained above in detail.

(Truncated Octahedron Three-Dimensional Puzzle)

The truncated octahedron three-dimensional puzzle of the presentinvention related to the present embodiment can be formed usingtwenty-four pieces which have a shape of an atom σ 1101 and twenty-fourwhich have a shape of the atom σ 1101 mirror image σ′ 1103 as explainedabove in detail.

(Rhombic Dodecahedron Three-Dimensional Puzzle)

The rhombic dodecahedron three-dimensional puzzle of the presentinvention related to the present embodiment can be formed usingninety-six pieces which have a shape of an atom σ 1101 and ninety-sixwhich have a shape of the atom σ 1101 mirror image σ′ 1103 as explainedabove in detail.

(Elongated Rhombic Dodecahedron Three-Dimensional Puzzle)

The elongated rhombic dodecahedron three-dimensional puzzle of thepresent invention related to the present embodiment can be formed usingone hundred and ninety-two pieces which have a shape of an atom σ 1101and one hundred and ninety-two which have a shape of the atom σ 1101mirror image σ′ 1103 as explained above in detail.

The three-dimensional puzzle of the present invention related to thepresent embodiment which has the shape of a Fedrov space filling solidcomprised of five types of polyhedron, namely a parallel hexagram,skewed hexagonal prism, truncated octahedron, rhombic dodecahedron andelongated rhombic dodecahedron, can be assembled from a piece which hasa shape of one atom σ 1101 and a piece which has the shape of one atomσ′ 1103 mirror image. As a result, it is possible to visually explainthe theorem 2 stated above and the three-dimensional puzzle of thepresent invention related to the present embodiment is excellent as aneducational material.

Fifth Embodiment (Elongated Rhombic Dodecahedron Including a SpaceFilling Solid)

The inventors discovered that the five types of Fedrov space fillingsolids each require a multiple of twenty-four c-squadrons, that is, amultiple of twenty-four atoms σ 1101 and σ′ 1103.

From the proof of the elongated rhombic dodecahedron in theorem 2 andFIG. 35, it is clear that the elongated rhombic dodecahedron 1500includes a rhombic dodecahedron 1300. In addition, from the proof of thecube 1200 and the rhombic dodecahedron 1300 in theorem 1, it was shownthat the rhombic dodecahedron 1300 includes the cube 1200. The rhombicdodecahedron 1300 is obtained by gluing the square bottom of thequadrangular pyramid 1201 to each face of the cube 1200. Therefore, theelongated rhombic dodecahedron 1500 includes the rhombic dodecahedron1300 and the rhombic dodecahedron 1300 includes the cube 1200.

The truncated octahedron 1600 includes the cube 1200 by a specialmethod. When atom σ 1101 and atom σ′ 1103 which are congruentpentahedrons shown in FIG. 38( b) are considered, a hexahedron 1701including seven vertices and eleven edges shown in FIG. 40 by gluing theatoms.

This hexahedron 1701 has a face comprised of three triangles (one rightisosceles triangle and two congruent right triangles), twoquadrilaterals and one pentagon. This polyhedron is called atripenquadron.

The polyhedron 1703 show in FIG. 41( a) is obtained when threetripenquadrons 1701 are glued along their right triangular faces so thatthey all meet at the vertex. The bottom face of this polyhedron 1703 isa regular hexahedron shown in FIG. 41( b).

When eight of these polyhedrons 1703 are glued along the right isoscelestriangular faces, a cube 1705 is obtained having six holes at the centerof each face as shown in FIG. 42( a). In order to examine the interiorof the cube 1705 with open hole, the cube is opened by a plane throughthe diagonals of two opposite faces which intersect these facesperpendicularly. When a diagonal cross section shown in FIG. 42( b) isviewed, it reveals that the cross section in the interior of thepolyhedron 1707 made by cutting the cube 1705 with open holes consist ofregular hexagons and that the hole is obtained by carving out half of atruncated octahedron 1603. FIG. 42( c) and FIG. 42( d) show a cube whichis the result of assembling the truncated octahedron and cube with openholes. It is clear that the truncated octahedron 1600 is included in thecube 1700 in a special way.

Finally, it is shown that a skewed hexagonal prism is included in anelongated rhombic dodecahedron in a particular way. The elongatedrhombic dodecahedron 1500 comprised of the rhombic dodecahedron 1300 andhelmet shaped polyhedron 1501 is again considered.

First, as is shown in FIG. 43( a), the rhombic dodecahedron 1300 isdivided into two congruent polyhedrons 1301 and only one of these isconsidered. Next, a part of the helmet shaped polyhedron 1501 shown inFIG. 43( b) is cut off as shown in FIG. 43( c). The right sidepolyhedron 1503 in FIG. 43( c) is V-shaped and consists of two skewedtriangular prisms 1505 and 1507 and the skewed triangle prism 1507 is amirror image of the skewed triangular prism 1505. Each skewed triangularprism is formed by gluing three sphenoids as shown in FIG. 36( a).

If the V-shaped polyhedron 1503 is glued to the lower half of therhombic dodecahedron 1301, the skewed hexagonal prism 1420 shown in FIG.43( d) and FIG. 43( e) is obtained. In this way, it is shown that theskewed hexagonal prism 1420 is included within the elongated rhombicdodecahedron 1500 in a special way.

The three-dimensional puzzle of the present invention related to thepresent embodiment which has the shape of a Fedrov space filling solidcomprised of five types of polyhedron, namely a cube, skewed hexagonalprism, rhombic dodecahedron, truncated octahedron, and elongated rhombicdodecahedron, can be assembled from a piece which has a shape of oneatom σ 1101 and a piece which has the shape of one atom σ′ 1103 of itsmirror image. In addition, it is possible to visually explain theinclusion of the cube, skewed hexagonal prism, rhombic dodecahedron andtruncated octahedron in the elongated rhombic dodecahedron and thethree-dimensional puzzle of the present invention related to the presentembodiment is excellent as an educational material.

(Truncated Octahedron Three-Dimensional Puzzle)

FIG. 44 shows the arrangement of a puzzle 1651 which is half of atruncated octahedron three-dimensional puzzle of the present inventionrelated to the present embodiment. (a) shows a side face view, (b) showsa side face view, (c) shows an upper face view and (d) shows a bottomface view. In addition, FIG. 45 shows the arrangement of a truncatedoctahedron three-dimensional puzzle 1650. (a) shows a side face view,(b) shows a side view and (c) shows a bottom face view. The truncatedoctahedron three-dimensional puzzle is formed using twenty-four pieceswhich have the shape of the atom σ 1101 and twenty-four pieces whichhave the shape of the mirror image σ′ 1103 of the atom σ 1101 asexplained in detail above.

(Cube Three-Dimensional Puzzle)

FIG. 46 shows an arrangement of a cube three-dimensional puzzle of thepresent invention related to the present embodiment. (a) shows thebottom face of a puzzle 1651 which is half of the truncated octahedronthree-dimensional puzzle, (b) shows a puzzle 1751 by a cross section ofa cube with open holes, (c) shows a puzzle 1753 of half of a cube inwhich the puzzle 1651 which is half of the truncated octahedronthree-dimensional puzzle is glued to puzzle 1751 formed by cutting thecube with open holes, (d) shows a cube three dimensional puzzle 1750.The cube three-dimensional puzzle can be formed using forty-eight pieceswhich have the shape of the atom σ 1101 and forty-eight pieces whichhave the shape of the mirror image σ′ 1103 of the atom δ 1101 asexplained in detail above.

(Rhombic Dodecahedron Three-Dimensional Puzzle)

FIG. 47 shows an arrangement of a rhombic dodecahedron three-dimensionalpuzzle of the present invention related to the present embodiment. (a)shows puzzle 1753 of half of a cube, (b) shows puzzle 1353 without thehalf of the cube from the puzzle 1351 which is half of a rhombicdodecahedron, (c) shows the puzzle 1351 which is half of a rhombicdodecahedron, (d) shows a side face view of rhombic dodecahedron puzzle1350, (e) shows an upper face view of the rhombic dodecahedron puzzle1350. The rhombic dodecahedron three-dimensional puzzle can be formedusing ninety-six pieces which have the shape of the atom σ 1101 andninety-six pieces which have the shape of the mirror image σ′ 1103 ofthe atom σ60 1101 as explained in detail above.

(Skewed Hexagonal Prism Three-Dimensional Puzzle)

FIG. 48 shows an arrangement of a skewed hexagonal prismthree-dimensional puzzle of the present invention related to the presentembodiment. (a) shows a side face view of puzzle 1553 cut from a helmetshaped polyhedron puzzle 1551, (b) shows a front face view of the puzzle1553, (c) shows a back face view of the puzzle 1553, (d) shows anexemplary view of the formation of a skewed hexagonal prism 1450 by thepuzzle 1553 which is cut away from the helmet shaped polyhedron puzzle1551 and a puzzle 1351 which is half of the rhombic dodecahedron 1350,(e) shows an upper face view of the skewed hexagonal prism 1450, (f)shows a side face view of the skewed hexagonal prism 1450. The skewedhexagonal prism 1450 three-dimensional puzzle can be formed usingseventy-two pieces which have the shape of the atom σ 1101 andseventy-two pieces which have the shape of the mirror image σ′ 1103 ofthe atom σ 1101 as explained in detail above.

(Elongated Rhombic Dodecahedron Three-Dimensional Puzzle)

FIG. 49 shows an arrangement of an elongated rhombic dodecahedronthree-dimensional puzzle of the present invention related to the presentembodiment. (a) shows a side face view of the helmet shaped polyhedronpuzzle 1551, (b) shows a front face view of the helmet shaped polyhedronpuzzle 1551, (c) shows a back view of the helmet shaped polyhedronpuzzle 1551, (d) shows an exemplary view of the formation of anelongated rhombic dodecahedron three-dimensional puzzle 1550 by thehelmet shaped polyhedron puzzle 1551 and the rhombic dodecahedron puzzle1350, (e) shows an upper face view of the elongated rhombic dodecahedronthree-dimensional puzzle 1550, (f) shows a side face view of theelongated rhombic dodecahedron three-dimensional puzzle 1550. Theelongated rhombic dodecahedron three-dimensional puzzle 1550 can beformed using one hundred and ninety-two which have the shape of the atomσ 1101 and one hundred and ninety-two which have the shape of the mirrorimage σ′ 1103 of the atom σ 1101 as explained in detail above.

The three-dimensional puzzle of the present invention related to thepresent embodiment which has the shape of a Fedrov space filling solidcomprised of five types of polyhedron, namely a cube, skewed hexagonalprism, rhombic dodecahedron, truncated octahedron and elongated rhombicdodecahedron, can be assembled from a piece which has a shape of oneatom σ 1101 and a piece which has the shape of one atom σ′ 1103 of itsmirror image. In addition, it is possible to visually explain thetheorem 2 stated above and the three-dimensional puzzle of the presentinvention related to the present embodiment is excellent as aneducational material.

Because the three-dimensional puzzle of the present invention is formedwith the minimum number of atoms (convex polyhedrons) for filing all ofthe regular polyhedrons, it is possible to form all of the regularpolyhedrons (regular tetrahedron, cube, regular octahedron, regulardodecahedron and regular icosahedron) even if the number of parts issmall. In addition, it is possible to visually prove the novel theoremdiscovered by the inventors and the present invention can be used as anexcellent educational material for explaining this theorem.

According to the three-dimensional puzzle of the present invention, byusing a different atom in addition to the minimum number of atoms forconstructing all of the regular polyhedrons, it is possible to form acube and regular dodecahedron without using one of the minimum number ofatoms. In addition, it is possible to visually explain the novel theoremdiscovered by the inventors, compare both puzzles by a transformationexample which uses a different atom, and the present invention can beused as an excellent educational material for explaining this theorem.

According to the three-dimensional puzzle of the present invention, evenif one of the minimum number of atoms for constructing all of theregular polyhedrons is not used, it is possible to form a regularpolyhedron, cube and regular dodecahedron by using a different atom. Inaddition, it is possible to visually explain the novel theoremdiscovered by the inventors, compare both puzzles by a transformationexample which uses a different atom, and the present invention can beused as an excellent educational material for explaining this theorem.

According to the present invention, it is possible to realize a Fedorovspace-filling solid and provide a three-dimensional puzzle which has notexisted until now.

1. A three-dimensional puzzle comprising: four types of convexpolyhedrons from which a regular tetrahedron, a cube, a regularoctahedron, a regular dodecahedron or a regular icosahedron are formed;wherein three types among the four types of convex polyhedrons eachhaving a pair of convex polyhedrons in a mirroring image relationship;the four type of convex polyhedrons are indivisible into two or morecongruent shaped polyhedrons; and the regular tetrahedron, the cube, theregular octahedron, the regular dodecahedron and the regular icosahedronare formed using only the four types of convex polyhedrons so that theinterior of the regular tetrahedron, the cube, the regular octahedron,the regular dodecahedron and the regular icosahedron are filled.
 2. Athree-dimensional puzzle comprising: five types of convex polyhedronsfrom which a regular tetrahedron, a cube, a regular octahedron, aregular dodecahedron or a regular icosahedron are formed; wherein fourtypes among the five types of convex polyhedrons each having a pair ofconvex polyhedrons in a mirroring image relationship; the five type ofconvex polyhedrons are indivisible into two or more congruent shapedpolyhedrons; and the regular tetrahedron, the cube, the regularoctahedron, the regular dodecahedron and the regular icosahedron areformed using only the five types of convex polyhedrons so that theinterior of the regular tetrahedron, the cube, the regular octahedron,the regular dodecahedron and the regular icosahedron are filled.